Arithmetic: percentage, ratio, averages, time & work

Answer: (d) 93.75%

The correct result is 4x and the mistaken one x/4. The result falls by 4x − x/4 = 15x/4, which is 15/16 of the correct result, i.e. 93.75%.

1. Take the number as 16. Correct result = 16 × 4 = 64; mistaken result = 16 ÷ 4 = 4.
2. Change = 64 − 4 = 60.
3. Percentage change = 60/64 × 100 = 93.75% (a fall).
4. Check in general: (4 − 1/4)/4 = 15/16 = 0.9375.

Remember · Pick a convenient number and measure the change against the correct result, which is the base.

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Answer: (b) 20%

20% failed in English and 30% in Hindi, with 15% failing both. So 5% failed only English and 15% failed only Hindi: 20% failed in exactly one subject.

1. Failed in English = 100 − 80 = 20%; failed in Hindi = 100 − 70 = 30%.
2. Failed in both = 15%.
3. Failed only in English = 20 − 15 = 5%; failed only in Hindi = 30 − 15 = 15%.
4. Failed in only one subject = 5 + 15 = 20%.
5. Check: passed in both = 100 − (20 + 30 − 15) = 65%, and 65 + 20 + 15 = 100.

Remember · Work with the failure sets: ‘only one’ = (A − both) + (B − both).

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Answer: (a) 30 years

n years ago the father was the son’s present age, so n is the gap between their ages. Using ‘father now = 4 × son n years ago’ gives father : son = 8 : 5, so with a total of 130 they are 80 and 50, a difference of 30 years.

1. Let the father’s age be F and the son’s S. ‘n years back I was as old as you are now’: F − n = S, so n = F − S.
2. Son’s age n years back = S − n = S − (F − S) = 2S − F.
3. F = 4(2S − F) → 5F = 8S → F : S = 8 : 5.
4. F + S = 130 → 13 parts = 130 → 1 part = 10, so F = 80 and S = 50.
5. Difference = 80 − 50 = 30 years.
6. Check: n = 30; the son was 20 thirty years ago, and 4 × 20 = 80.

Remember · ‘I was your present age n years ago’ means n equals the age gap; turn the rest into a ratio.

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Answer: (b) 42 kg

Four women weigh 4 × 63 = 252 kg, and six boys weigh the same. So the average boy weighs 252 ÷ 6 = 42 kg.

1. Weight of 4 women = 4 × 63 = 252 kg.
2. Weight of 6 boys = weight of 4 women = 252 kg.
3. Average weight of a boy = 252 ÷ 6 = 42 kg.
4. Check: 6 × 42 = 252 = 4 × 63.

Remember · In a chain of equal totals, use only the two links the question needs and ignore the rest.

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Answer: (d) ₹1,350

Take Q’s cost as 100: R costs 120 and P costs 150, a total of 370 units. As 370 units = ₹3,330, one unit = ₹9, so P costs 150 × 9 = ₹1,350.

1. Let Q = 100. R is 20% more: 120. P is 25% more than R: 150.
2. Total = 100 + 120 + 150 = 370 units = ₹3,330, so 1 unit = ₹9.
3. P = 150 × 9 = ₹1,350.
4. Check: Q = ₹900, R = ₹1,080, P = ₹1,350, and 900 + 1,080 + 1,350 = 3,330.

Remember · Chain the percentages from the cheapest item taken as 100, then scale to the given total.

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Answer: (c) Both 1 and 2

4O + 6M + 8A = 2(O + 2M + 5A) simplifies to O + M = A, which is statement 2. Statement 1 then follows, because 3O + 5M + 9A differs from 4O + 6M + 8A by A − O − M = 0.

1. Let the prices be O, M and A. Then 4O + 6M + 8A = 2O + 4M + 10A.
2. So 2O + 2M = 2A, i.e. O + M = A. Statement 2 is correct.
3. (3O + 5M + 9A) − (4O + 6M + 8A) = A − O − M = 0. Statement 1 is correct.
4. Check with O = 1, M = 2, A = 3: 4 + 12 + 24 = 40 = 2 × (1 + 4 + 15), and 3 + 10 + 27 = 40 as well.

• ✓ 1. The two totals differ by A − O − M, which is 0 because O + M = A.
• ✓ 2. Simplifying the given condition gives exactly O + M = A.

Remember · Simplify the given condition first; then test each statement by subtracting one side from the other.

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Answer: (a) 1 only

Take the work as 48 units: A does 6, B 3 and C 4 units a day, so each 3-day round finishes 13 units. After 9 days 39 units are done; A adds 6 on day 10 and B completes the last 3 on day 11. Counting from Monday, day 11 is a Thursday.

1. Total work = LCM(8, 16, 12) = 48 units. A does 6, B does 3 and C does 4 units a day.
2. One round (A, B, C) = 6 + 3 + 4 = 13 units in 3 days. After 3 rounds (9 days): 39 units, so 9 units remain.
3. Day 10 (A): 39 + 6 = 45 units. Day 11 (B): 45 + 3 = 48 units — the work is complete.
4. Day 1 is Monday, so day 8 is Monday again and day 11 is Thursday.
5. Statement 1 is correct; statement 2 is not, because the work takes 11 days.

• ✓ 1. The work ends on day 11, and day 11 counted from a Monday is a Thursday.
• ✗ 2. After 10 days only 45 of the 48 units are done; B needs day 11 to finish.

Remember · Take the LCM of days as total work, add round by round, then day by day; convert day numbers to weekdays using remainders by 7.

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Answer: (c) R < S

At the same rate, half-yearly compounding earns interest on interest and so grows faster than annual compounding. To reach the same amount Q in one year, the half-yearly rate R must therefore be smaller than the annual rate S.

1. Half-yearly at R% a year: Q = P(1 + R/200)². Annually at S%: Q = P(1 + S/100).
2. So 1 + S/100 = (1 + R/200)² = 1 + R/100 + R²/40000.
3. S = R + R²/400, which is greater than R for any positive rate.
4. Example: R = 10 gives (1.05)² = 1.1025, so S = 10.25. Hence R < S.
5. Option (d), R ≤ S, is not false as a statement, but equality never occurs for a positive rate; the exact relation is R < S.

Remember · More frequent compounding at the same rate gives more; to give the same amount, the more frequent rate must be lower.

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Answer: (d) 1, 2 and 3

Let a, b, c be the numbers of ₹50, ₹20 and ₹10 notes. The two totals reduce to 4a + b = 134, and none of the three statements gives whole, non-negative numbers of notes, so all three are incorrect.

1. Let the numbers of ₹50, ₹20 and ₹10 notes be a, b and c. Then a + b + c = 50 and 50a + 20b + 10c = 1840, that is 5a + 2b + c = 184.
2. Subtract the first equation from the second: 4a + b = 134.
3. Statement 1: a = 25 gives b = 134 − 100 = 34, and then c = 50 − 25 − 34 = −9. Impossible.
4. Statement 2: b = 35 gives 4a = 99, which is not a whole number. Impossible.
5. Statement 3: c = 20 gives a + b = 30; with 4a + b = 134 this means 3a = 104, not a whole number. Impossible.
6. Check: a valid payment does exist, e.g. a = 30, b = 14, c = 6 — that is 50 notes and 1500 + 280 + 60 = ₹1,840 — so the bill itself is fine; only the three statements fail.

• ✗ 1. a = 25 forces c = −9, a negative number of ₹10 notes.
• ✗ 2. b = 35 forces 4a = 99, so a is not a whole number.
• ✗ 3. c = 20 forces 3a = 104, so a is not a whole number.

Remember · With a count total and a value total, subtract to get one clean equation, then test each statement for whole, non-negative answers.

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Answer: (b) 3

By the time X finishes 3 km, Y has run 2 km, so X has gained 1000 m — three full laps of 300 m and a little more. Each full lap gained is one pass, so X passes Y 3 times.

1. The race ends when X, the faster runner, completes 3 km = 3000 m.
2. Speeds are 3 : 2, so in that time Y runs 2/3 × 3000 = 2000 m.
3. X gains 3000 − 2000 = 1000 m on Y.
4. X passes Y each time the lead reaches a whole lap: at 300 m, 600 m and 900 m. 1000 ÷ 300 = 3.33, so there are 3 passes.

Remember · On a circular track, passes = lead gained by the faster runner ÷ track length, counting whole laps only.

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Answer: (b) 10%

A 25% rise followed by a 20% fall brings the price exactly back to where it started (1.25 × 0.8 = 1). Only the last 10% rise remains.

1. Let the price be ₹100. After a 25% increase it is ₹125.
2. A 20% decrease: 125 × 0.8 = ₹100.
3. A 10% increase: 100 × 1.1 = ₹110.
4. The resultant increase is ₹10 on ₹100, i.e. 10%.

Remember · Successive percentage changes multiply: +25% then −20% cancel, because 5/4 × 4/5 = 1.

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Answer: (b) 40.4 m

With plants at both ends of the road, 101 plants make 100 equal gaps along 1010 m, so each gap is 10.1 m. Five consecutive plants span 4 gaps: 4 × 10.1 = 40.4 m.

1. 101 plants in a line from one end of the road to the other leave 101 − 1 = 100 gaps.
2. Road length = 1.01 km = 1010 m, so each gap = 1010 ÷ 100 = 10.1 m.
3. 5 consecutive plants have 4 gaps between them: 4 × 10.1 = 40.4 m.
4. Trap: dividing 1010 by 101 (counting plants instead of gaps) gives 10 m and the wrong answer 40 m.

Remember · n objects in a line make n − 1 gaps; k consecutive objects span k − 1 gaps.

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Answer: (d) 52

Work backwards from D’s 10 coins. Each giver handed over half his coins plus 2, so he had 2 × (coins given − 2): C had 16, B had 28 and A had 52.

1. D now has 10 coins (the smallest two-digit number), all received from C.
2. C gave half his coins + 2 = 10, so half = 8 and C had 16 — all received from B.
3. B gave half + 2 = 16, so half = 14 and B had 28 — all received from A.
4. A gave half + 2 = 28, so half = 26 and A had 52.
5. Check: A (52) gives 26 + 2 = 28 to B; B gives 14 + 2 = 16 to C; C gives 8 + 2 = 10 to D.

Remember · For ‘gives half and k more’ chains, reverse each step: original = 2 × (amount given − k).

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Answer: (d) 694861x

Convert each unit to seconds: a week is 604800 s, a day 86400 s, an hour 3600 s, a minute 60 s. Adding these and the lone second gives 694861 seconds for every x.

1. 1 week = 7 × 24 × 3600 = 604800 s; 1 day = 24 × 3600 = 86400 s; 1 hour = 3600 s; 1 minute = 60 s.
2. Total = (604800 + 86400 + 3600 + 60 + 1)x = 694861x seconds.
3. Check: the x seconds make the coefficient end in 1, and one week alone is 604800x, so only 694861x fits.

Remember · Use the last digit as a filter: the extra ‘x seconds’ makes the total end in 1.

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Answer: (a) ₹2,600

Statements 2 and 3 give P = 13 and Q = 8; then X = 8, Z = 16 and Y = 20. The five bought 65 notebooks, which cost 65 × ₹40 = ₹2,600.

1. From 2 and 3: P + Q = 21 and P − Q = 5, so P = 13 and Q = 8.
2. From 5: X = 13 − 5 = 8. From 1: Z = 8 + 8 = 16. From 4: Y = 28 − 8 = 20.
3. Total notebooks = (P + Q) + (X + Y) + Z = 21 + 28 + 16 = 65.
4. Cost = 65 × ₹40 = ₹2,600.

Remember · Start with the pair of equations that closes on its own (sum and difference), then chain outwards.

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Answer: (d) 5 minutes fast

Of the 90 minutes he was away by the home clock, 65 were spent driving. The return took 0.625 of the outward time, so the outward drive took 40 minutes and he reached the village at 15:10 home time — the village clock, showing 15:15, is 5 minutes fast.

1. By the home clock he was away from 14:30 to 16:00 = 90 minutes; 25 minutes were spent in the village, so driving took 65 minutes.
2. Let the outward drive take t minutes. The return route is 1.25 times as long at twice the speed, so it takes 1.25 ÷ 2 × t = 0.625t.
3. t + 0.625t = 65, so 1.625t = 65 and t = 40 minutes.
4. He reached the village at 14:30 + 40 min = 15:10 by the home clock, when the village clock showed 15:15.
5. So the village clock is 5 minutes fast.
6. Check: he leaves the village at 15:35 (home time); the return takes 0.625 × 40 = 25 minutes, reaching home at 16:00.

Remember · Use only one clock for elapsed time; compare the other clock’s reading with the true time at that moment.

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Answer: (d) Data is inadequate to draw any conclusion

The answer depends on how a man’s work rate compares with a woman’s, and that is not given. Equal rates give exactly 30 days, faster men give more than 30 and faster women fewer — so the data are inadequate.

1. Let a man do m units and a woman w units of work a day. Work = 30 × (24m + 12w).
2. Days for 12 men and 24 women = 30(24m + 12w)/(12m + 24w) = 30(2m + w)/(m + 2w).
3. If m = w this is 30 days; if m > w it is more than 30; if m < w it is less than 30.
4. The relation between m and w is not given, so no conclusion can be drawn. Option (c) fails because it leaves out the possibility of exactly 30 days.

Remember · In time-and-work items with two kinds of workers, you need their relative efficiency before you can swap one for the other.

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Answer: (a) z < x < y

The first condition gives 0.7x = 0.65y, so y = 14x/13, just above x. The second gives 0.6x = 0.65z, so z = 12x/13, just below x. Hence z < x < y.

1. 0.7x + y = 1.65y, so 0.7x = 0.65y and y = 70x/65 = 14x/13.
2. 0.6x + z = 1.65z, so 0.6x = 0.65z and z = 60x/65 = 12x/13.
3. For positive x, 12x/13 < x < 14x/13, so z < x < y.
4. Check with x = 13: y = 14 and z = 12; 9.1 + 14 = 23.1 = 1.65 × 14 and 7.8 + 12 = 19.8 = 1.65 × 12.

Remember · Turn each percentage condition into ‘part of x = part of y’ and compare the resulting multiples of x.

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Answer: (a) 15000

X got 56% and Y 44% of the votes cast, so the winning margin is 12% of the votes cast: 1440 votes. That makes 12000 votes cast, which is 80% of the list — so the list has 15000 voters.

1. With two candidates and no invalid votes, Y got 100% − 56% = 44% of the votes cast.
2. Margin = 56% − 44% = 12% of the votes cast = 1440, so votes cast = 1440 ÷ 0.12 = 12000.
3. Votes cast are 80% of the voters list, so total voters = 12000 ÷ 0.8 = 15000.
4. Check: 12000 is 80% of 15000; X gets 6720, Y gets 5280, and 6720 − 5280 = 1440.

Remember · In two-candidate elections, the margin is (winner% − loser%) of the votes cast; convert back to the full list last.

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Answer: (a) m = n

After the two transfers each container again holds 100 ml. Whatever water went into X must be matched by the same amount of milk left in Y, so X’s milk share equals Y’s water share: m = n = 5/6.

1. After the first transfer: X has 80 ml milk; Y has 100 ml water + 20 ml milk = 120 ml.
2. 20 ml of Y’s mixture carries 20 × 20/120 = 10/3 ml milk and 20 × 100/120 = 50/3 ml water back to X.
3. X: milk = 80 + 10/3 = 250/3 ml out of 100 ml, so m = 5/6.
4. Y: water = 100 − 50/3 = 250/3 ml out of 100 ml, so n = 5/6.
5. So m = n.

Remember · Back-and-forth transfers of equal volume: the milk in the water jar always equals the water in the milk jar.

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Answer: (c) 41 kg

A + B + C = 120 and B + D + E = 126. Since F weighs the same as B, the six weights total 120 + 126 = 246 kg, and the average is 41 kg even though B itself is unknown.

1. A + B + C = 3 × 40 = 120 kg and B + D + E = 3 × 42 = 126 kg.
2. Since F = B, A + B + C + D + E + F = (A + B + C) + (F + D + E) = 120 + (B + D + E) = 120 + 126 = 246 kg.
3. Average = 246 ÷ 6 = 41 kg.

Remember · Before answering ‘cannot be determined’, check whether the unknown cancels — here F = B fills the gap exactly.

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Answer: (b) 20th May

The two formulas never meet within the first 100 days, after which petrol stays at ₹90. Diesel reaches ₹90 on day 140, and day 140 of 2021 (not a leap year) is 20 May.

1. While both formulas run (days 1–100): 80 + 0.1m = 69 + 0.15m gives 0.05m = 11, so m = 220 — outside 1 to 100, so no match in this period.
2. From day 100 onwards petrol stays at 80 + 0.1 × 100 = ₹90.
3. Diesel reaches ₹90 when 69 + 0.15n = 90, i.e. 0.15n = 21, so n = 140.
4. Days up to 30 April: 31 + 28 + 31 + 30 = 120, so day 140 is the 20th day of May — 20 May.
5. Check: on day 140 diesel = 69 + 0.15 × 140 = 69 + 21 = ₹90, equal to petrol.

Remember · For prices defined piece by piece, test each piece separately; then turn the day number into a date month by month.

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Answer: (d) P = 50 (2)^(n/12)

Doubling every 12 years is exponential growth: after n years the population has doubled n/12 times, so P = 50 × 2^(n/12). Putting n = 0 and n = 12 (giving 50 and 100) confirms it and rules out the other options.

1. Doubling every 12 years means multiplying by 2 once for each 12-year block.
2. In n years there are n/12 such blocks, so P = 50 × 2^(n/12).
3. Check: n = 0 gives 50; n = 12 gives 100; n = 24 gives 200 — doubling every 12 years.
4. Options (a) and (b) add a fixed number each year (linear growth); (c) gives 50 × 2¹² at n = 1, far too fast.

Remember · Test formula options by plugging in simple values — the start (n = 0) and one full period.

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Answer: (d) 20%

Take 100 students: 60 Indian and 40 foreign, with 50 girls. Indian girls are 18, so foreign girls are 32 and foreign boys only 8 — that is 8 out of 40 foreign students, or 20%.

1. Take 100 students: 60 Indian, 40 foreign; 50 girls in all.
2. Indian girls = 30% of 60 = 18.
3. Foreign girls = 50 − 18 = 32, so foreign boys = 40 − 32 = 8.
4. Share of boys among foreign students = 8/40 = 20%.
5. Check: boys = 50; Indian boys 60 − 18 = 42 plus foreign boys 8 = 50.

Remember · Assume a total of 100 and fill a 2 × 2 table; then divide by the base the question names (here, foreign students).

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Answer: (b) 5

Why not the tempting option · UPSC's key is 5. A very literal reading of 'previous height' lets the ball rise once more, to 49.15 cm, and touch the ground a sixth time; but that final contact is the ball coming to rest, not a hit before it stops bouncing, so 5 remains the better count. In the exam, stop at the first height that breaks the threshold and do not add the final landing.

Each bounce reaches 4/5 of the previous height: 150 → 120 → 96 → 76.8 → 61.44 cm. The next rise would be only 49.15 cm, below the 50 cm limit, so the ball does not bounce again; by then it has hit the ground 5 times.

1. Heights in cm: dropped from 150, the ball rises to 4/5 of each previous height — 120, 96, 76.8, 61.44.
2. Every rise is preceded by a hit: hits after falling from 150, 120, 96, 76.8 and 61.44 cm — that is 5 hits.
3. The rise after the 5th hit would be 4/5 × 61.44 = 49.152 cm, below 50 cm, so the ball does not bounce further.
4. Number of times the ball hits the ground before it stops bouncing = 5.

Remember · List the geometric sequence of heights, stop at the first term that breaks the threshold, and count the hits up to that point; do not add an extra landing after the ball has stopped bouncing.

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Answer: (c) Both 1 and 2

Adding the hourly distances, the two close the 15 km gap exactly at the end of the 5th hour. X walks 5 × 1.5 = 7.5 km and Y walks 1 + 1.25 + 1.5 + 1.75 + 2 = 7.5 km, so they meet midway — both statements hold.

1. Distance closed in each hour: 1.5 + 1 = 2.5, then 1.5 + 1.25 = 2.75, then 3, 3.25, 3.5 km.
2. Running total: 2.5, 5.25, 8.25, 11.5, 15 km — they meet at the end of the 5th hour.
3. X covers 5 × 1.5 = 7.5 km; Y covers 1 + 1.25 + 1.5 + 1.75 + 2 = 7.5 km.
4. Each covers half of 15 km, so they meet midway.
5. Check: Y's average speed over 5 hours is 7.5 ÷ 5 = 1.5 km/hr, equal to X's speed — hence midway.

• ✓ 1. The combined distance reaches 15 km exactly after 5 hours.
• ✓ 2. Both walk 7.5 km in those 5 hours, so the meeting point is the midpoint.

Remember · With speeds changing hour by hour, tabulate the hours; for an arithmetic progression of speeds, the average speed gives a quick check.

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Answer: (b) 3

With equal maximum marks, the overall 60% equals the average of the six papers' shares, and the average of 5 to 10 is 7.5. Only the papers with shares 5, 6 and 7 fall below that level — 3 papers.

1. Let the marks be 5k, 6k, 7k, 8k, 9k and 10k; the total is 45k.
2. Since every paper has the same maximum, the overall 60% equals the average paper score: 45k ÷ 6 = 7.5k is 60% of the maximum.
3. Papers below 7.5k: 5k, 6k and 7k — 3 papers.
4. Check: if the maximum is 12.5k, the marks are 40%, 48%, 56%, 64%, 72%, 80% — three below 60%.

Remember · When maximum marks are equal, the overall percentage is the average of the paper percentages; compare each paper with the mean.

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Answer: (d) 30 or more

There are only 40 foreigners, so at most 40 of the 70 English speakers can be foreigners. At least 30 must be Indians, and the number could be higher, so the answer is '30 or more'.

1. Foreigners = 120 − 80 = 40.
2. Even if all 40 foreigners speak English, at least 70 − 40 = 30 English speakers are Indians.
3. If all 70 English speakers were Indians, the number would be 70; the data do not fix it.
4. So the number of Indians who speak English is 30 or more.

Remember · Minimum overlap of two groups = (size of A + size of B) − total; give a range when the data do not fix an exact value.

Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) 1 only

Every shifted student scored at most 21, below Class-B's lowest score of 22, so Class-B's average must fall. Class-A's average can rise or fall depending on which four students leave, so statement 2 is not certain.

1. Each shifted student scored at most 21; every Class-B student scored at least 22.
2. Adding scores lower than all existing scores pulls Class-B's average down — statement 1 is certain.
3. Class-A's new average depends on whether the four who left were above or below Class-A's average.
4. If the four scored 21 each, Class-A's average falls; if they scored 17 each, it rises — so statement 2 is not definite.

• ✓ 1. New members all score below Class-B's minimum, so its average must decrease.
• ✗ 2. Removing students raises the average only if they were below it; the data do not say which four moved.

Remember · Adding values below every existing value lowers the mean; removing values changes it only relative to the current mean.

Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) The price of a pen is two times the price of a pencil

Both spent the same, so 3 pens + 5 pencils cost as much as 2 pens + 7 pencils. Cancelling the common items leaves 1 pen = 2 pencils.

1. Equal spending: 3 pens + 5 pencils = 2 pens + 7 pencils.
2. Remove 2 pens and 5 pencils from both sides: 1 pen = 2 pencils.
3. So a pen costs twice as much as a pencil.

Remember · With equal totals, cancel what both sides share — the leftovers must balance.

Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error